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Section 4-10 : Ratio Test

In this section we are going to take a look at a test that we can use to see if a series is absolutely convergent or not. Recall that if a series is absolutely convergent then we will also know that it’s convergent and so we will often use it to simply determine the convergence of a series.

Before proceeding with the test let’s do a quick reminder of factorials. This test will be particularly useful for series that contain factorials (and we will see some in the applications) so let’s make sure we can deal with them before we run into them in an example.

If \(n\) is an integer such that \(n \ge 0\) then \(n\) factorial is defined as,

So, \(L < 1\) and so by the Ratio Test the series converges absolutely and hence will converge.

As seen in the previous example there is usually a lot of canceling that will happen in these. Make sure that you do this canceling. If you don’t do this kind of canceling it can make the limit fairly difficult.

Example 2 Determine if the following series is convergent or divergent.
\[\sum\limits_{n = 0}^\infty {\frac{{n!}}{{{5^n}}}} \]

Show Solution

Now that we’ve worked one in detail we won’t go into quite the detail with the rest of these. Here is the limit.

In order to do this limit we will need to eliminate the factorials. We simply can’t do the limit with the factorials in it. To eliminate the factorials we will recall from our discussion on factorials above that we can always “strip out” terms from a factorial. If we do that with the numerator (in this case because it’s the larger of the two) we get,

In the previous example the absolute value bars were required to get the correct answer. If we hadn’t used them we would have gotten \(L = - \frac{9}{2} < 1\) which would have implied a convergent series!

Now, let’s take a look at a couple of examples to see what happens when we get\(L = 1\). Recall that the ratio test will not tell us anything about the convergence of these series. In both of these examples we will first verify that we get \(L = 1\) and then use other tests to determine the convergence.

So, as implied earlier we get \(L = 1\) which means the ratio test is no good for determining the convergence of this series. We will need to resort to another test for this series. This series is an alternating series and so let’s check the two conditions from that test.

So, as we saw in the previous two examples if we get \(L = 1\) from the ratio test the series can be either convergent or divergent.

There is one more thing that we should note about the ratio test before we move onto the next section. The last series was a polynomial divided by a polynomial and we saw that we got \(L = 1\) from the ratio test. This will always happen with rational expression involving only polynomials or polynomials under radicals. So, in the future it isn’t even worth it to try the ratio test on these kinds of problems since we now know that we will get \(L = 1\).

Also, in the second to last example we saw an example of an alternating series in which the positive term was a rational expression involving polynomials and again we will always get \(L = 1\) in these cases.

Let’s close the section out with a proof of the Ratio Test.

Proof of Ratio Test

First note that we can assume without loss of generality that the series will start at \(n = 1\) as we’ve done for all our series test proofs.

Let’s start off the proof here by assuming that \(L < 1\) and we’ll need to show that \(\sum {{a_n}} \) is absolutely convergent. To do this let’s first note that because \(L < 1\) there is some number \(r\) such that \(L < r < 1\).

So, for \(k = 1,2,3, \ldots \) we have \(\left| {{a_{N + k}}} \right| < {r^k}\left| {{a_N}} \right|\). Just why is this important? Well we can now look at the following series.

\[\sum\limits_{k = 0}^\infty {\left| {{a_N}} \right|{r^k}} \]

This is a geometric series and because \(0 < r < 1\) we in fact know that it is a convergent series. Also because \(\left| {{a_{N + k}}} \right| < {r^k}\left| {{a_N}} \right|\) by the Comparison test the series

we know that \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) is also convergent since the first term on the right is a finite sum of finite terms and hence finite. Therefore \(\sum\limits_{n = 1}^\infty {{a_n}} \) is absolutely convergent.

Next, we need to assume that \(L > 1\) and we’ll need to show that \(\sum {{a_n}} \) is divergent. Recalling that,

because the terms are getting larger and guaranteed to not be negative. This in turn means that,

\[\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0\]

Therefore, by the Divergence Test \(\sum {{a_n}} \) is divergent.

Finally, we need to assume that \(L = 1\) and show that we could get a series that has any of the three possibilities. To do this we just need a series for each case. We’ll leave the details of checking to you but all three of the following series have \(L = 1\) and each one exhibits one of the possibilities.